Using force equations and kinematic equations

A 1.0kg wood block is launched up a wooden ramp that is inclined at a 38 degree angle. The block's initial speed is 14 m/s. (Use uk = .20 for the coefficient of kinetic friction for wood on wood.) a.) What vertical height does the block reach above its starting point? b.) What speed does it have when it slides back down to its starting point?

2. Relevant equations

F = ma
Frictional force = uk(n)
kinematic equations

3. The attempt at a solution
I broke the weight into its components. Then I found the forces for each direction. I then solved for the normal force, and got 59.249N. I then substituted that in for the equation in the x-direction to find the acceleration. I got -67.179 m/s^2. I then tried using Vxf^2 = Vxi^2 + 2(ax)(delta x).

A 1.0kg wood block is launched up a wooden ramp that is inclined at a 38 degree angle. The block's initial speed is 14 m/s. (Use uk = .20 for the coefficient of kinetic friction for wood on wood.) a.) What vertical height does the block reach above its starting point? b.) What speed does it have when it slides back down to its starting point?

2. Relevant equations

F = ma
Frictional force = uk(n)
kinematic equations

3. The attempt at a solution
I broke the weight into its components. Then I found the forces for each direction. I then solved for the normal force, and got 59.249N. I then substituted that in for the equation in the x-direction to find the acceleration. I got -67.179 m/s^2. I then tried using Vxf^2 = Vxi^2 + 2(ax)(delta x).

I cannot see friction in your equations(or method), and normal force cannot be 59 N. It is much smaller